Posted by **Claire** on Tuesday, February 15, 2011 at 3:01am.

Hi can some one help, explain the following please. I seem to have issues with this:

1- Find the corresponding particular solution (in implicit form) that satisfies the initial; condition y = when x = 0.

2- Find the explicit form of this particular solution.

3- What is the value of y given by this particular solution when x =1? Give your answer to four decimal places.

regards Claire

- differential equations -
**MathMate**, Tuesday, February 15, 2011 at 9:52am
Take the example of a differential equation:

Since the differential equation in question has not been posted, I will work with an example.

y"-2y'+2y=e^{2x}.....(1)

The solution yh to the homogeneous equation

y"-2y'+2y=0 .... (2)

is

yh=C1*e^{x}+C2*x*e^{x}

Note the two integration constants.

To obtain the general solution to equation (1), we must add to yh, solution of (2) a particular solution that satisfies (1). This is called a particular solution, yp.

For this particular problem,

yp=x²e^{x}

which can be obtained by the method of variation of parameters.

Therefore the general solution of (1) is the sum of yh and yp, i.e.

y = C1*e^{x} + C2*x*e^{x} + x²e^{x} ...(general solution)

However, y still contains the constants C1 and C2 to be determined. They can be found by the initial conditions, such as y(0)=0 and y'(0)=1.

Substitute (0,0) into the general solution, we obtain an equation in terms of C1 and C2.

Differentiate the general solution once and substitute x=0, y'(0)=1, we get another equation. From these equations, we can solve for C1 and C2 which is the solution to the problem.

The above general solution is explicit because y, and only y, appears only on the left hand side.

A solution of the form

xy²+x²=2y²

is called an implicit solution because y appears more than once in an equation.

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